Theorem - Let $X$ be an Hilbert space and $A \subseteq X$ a closed subspace. Then the orthogonal complement of $A$ , denoted $A^{\perp}$ , is a topological complement of $A$ . That means $A^{\perp}$ is closed and

Proof :

• $A^{\perp}$ is closed :

  This follows easily from the continuity of the inner product. If a sequence $(x_n)$ of elements in $A^{\perp}$ converges to an element $x_0 \in X$ , then

     for every
  which implies that $x_0 \in A^{\perp}$ .
• $X=A \oplus A^{\perp}$ :

  Since $X$ is complete and $A$ is closed, $A$ is a complete subspace of $X$ . Therefore, for every $x \in X$ , there exists a best approximation of $x$ in $A$ , which we denote by $a_0 \in A$ , that satisfies $x-a_0 \in A^{\perp}$ (see this entry).

  This allows one to write $x$ as a sum of elements in $A$ and $A^{\perp}$

  
  which proves that
  

  Moreover, it is easy to see that

  
  since if $y \in A \cap A^{\perp}$ then $\langle y, y \rangle = 0$ , which means $y=0$ .

  We conclude that $X=A \oplus A^{\perp}$ . $\square$